Problem: The polynomial $p(x)=x^3-21x-20$ has a known factor of $(x-5)$. Rewrite $p(x)$ as a product of linear factors. $p(x)=$
Solution: We know $(x-5)$ is a factor of $p(x)$. This means that $p(x)=(x-5)\cdot q(x)$ for some polynomial $q(x)$. We can find $q(x)$ using polynomial division, and then we can factor $q(x)$. This way, we will be able to rewrite $p(x)$ as a product of linear factors. Dividing $p(x)$ by $(x-5)$ Notice that $p(x)$ is missing a $2^{\text{nd}}$ degree term. Let's add it as $0x^2$. $\begin{array}{r} x^2+\phantom{1}5x+\phantom{1}4 \\ x-5|\overline{x^3+0x^2-21x-20} \\ \mathllap{-(}\underline{x^3-5x^2\phantom{-21x-20}\rlap)} \\ 5x^2-21x-20 \\ \mathllap{-(}\underline{5x^2-25x\phantom{-20}\rlap)} \\ 4x-20 \\ \mathllap{-(}\underline{4x-20\rlap)} \\ 0 \end{array}$ We find that $q(x)=x^2+5x+4$. Factoring $q(x)$ We can factor $x^2+{5}x+{4}$ as $(x+m)(x+n)$ where $m+n={5}$ and $m\cdot n={4}$. Such numbers are $4$ and $1$, so the factored expression is $(x+4)(x+1)$. Putting it all together $\begin{aligned} p(x)&=x^3-21x-20 \\\\ &=(x-5)(x^2+5x+4) \\\\ &=(x-5)(x+4)(x+1) \end{aligned}$